x^3 −1 = (x−1) (x^2 +x+1) ⇒ x = 1 and x = ((−1±(√3) i)/2) ⇒ w = −(1/2) + (((√3)i )/2) and w^2 = −(1/2) − ((√3)/2) i ⇒ w^3 = 1 similarly x^3 + 1= 0 ⇒ x = −1 and x =−w^2 and x = −w


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Question Number 191640 by vishal1234 last updated on 28/Apr/23

$x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ $\Rightarrow x = 1 a n d x = \frac{- 1 \pm \sqrt{3} i}{2}$ $\Rightarrow w = - \frac{1}{2} + \frac{\sqrt{3} i}{2} a n d w^{2} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$ $\Rightarrow w^{3} = 1$ $s i m i l a r l y x^{3} + 1 = 0$ $\Rightarrow x = - 1 a n d x = - w^{2} a n d x = - w$
Commented by a.lgnaoui last updated on 28/Apr/23

$w = - \frac{1}{2} + \frac{\sqrt{3}}{2} i = j (j = e^{i \frac{2 π}{3}})$
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